Problem: Find the sum of the geometric series $1 -3 + 3^2 -3^3 +... -3^{29}$ Choose 1 answer: Choose 1 answer: (Choice A) A $-1.03\cdot10^{14}$ (Choice B) B $-6.86\cdot10^{13}$ (Choice C) C $ -5.15\cdot10^{13} $ (Choice D) D $1.72\cdot10^{13}$
Solution: Getting started We're dealing with a geometric series because each term is multiplied by $-3$ to get the next term. We need a formula to compute the sum of the terms. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The first term $(a_1 = {1})$ is given in the question. The number of terms $n$ is ${30}$ because there are ${30}$ numbers from $0$ to $29$. [Where do the 0 and 29 come from?] The common ratio $r$ is ${-3}$ because each term is multiplied by ${-3}$ to get the next term. [How did we find the common ratio r?] Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{30}}&=\dfrac{{1}(1-\left({-3}\right)^{{30}})}{1-\left({-3}\right)} \\\\ S_{{30}}&=\dfrac14(1-\left({-3}\right)^{{30}}) \\\\ S_{{{30}}} &\approx -5.15 \cdot10^{13} \end{aligned}$ The answer $ -5.15 \cdot10^{13} $